\newproblem{lay:2_1_17}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.17}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Ana Peña Gil, Jan. 19th, 2014} \\}{}

  % Problem statement
  If $A=\begin{pmatrix}1 & -3 \\ -3 & 5\end{pmatrix}$ and $AB=\begin{pmatrix}-3 & -11 \\ 1 & 17\end{pmatrix}$, determine the first and second columns of B. \\
}{
  % Solution
	We search for a matrix $B=\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{pmatrix}$ such that
	\begin{center}
	   $AB=\begin{pmatrix}1 & -3 \\ -3 & 5\end{pmatrix}\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{pmatrix}=
		     \begin{pmatrix}  b_{11}-3b_{21} & b_{12}-3b_{22} \\ 5b_{21}-3b_{11}& 5b_{22}-3b_{12}\end{pmatrix}=\begin{pmatrix}-3 & -11 \\ 1 & 17\end{pmatrix}$
	\end{center}
	This matrix equation gives us 4 equations
	\begin{center}
	   $b_{11}-3b_{21}=-3$ \\
		 $b_{12}-3b_{22}=-11$ \\
		 $5b_{21}-3b_{11}=1$ \\
		 $5b_{22}-3b_{12}=17$
	\end{center}
	The augmented matrix of this equation system is
	\begin{center}
	   $\left(\begin{array}{rrrr|r} 1 & 0 & -3 & 0 & -3 \\ 0 & 1 & 0 & -3 & -11 \\ -3 & 0 & 5 & 0 & 1 \\ 0 & -3 & 0 & 5 & 17 \end{array}\right) \sim 
		  \left(\begin{array}{rrrr|r} 1 & 0 & -3 & 0 & -3 \\ 0 & 1 & 0 & -3 & -11 \\ 0 & 0 & -4 & 0 & -8 \\ 0 & 0 & 0 & -4 & -16 \end{array}\right)$
	\end{center}
	Consequently, $b_{11}=3$, $b_{12}=1$, $b_{21}=2$ and $b_{22}=4$. The first and second columns of B are:
	\begin{center}
		$B=\begin{pmatrix}3 & 1 \\ 2 & 4\end{pmatrix}$
	\end{center}
}
\useproblem{lay:2_1_17}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
